A 182 gram sample of gold at some temperature is added to 22.1 grams of water. The initial water temperature is 25.0 °C, the final temperature is 27.5°C. If the specific heat capacity of gold is 0.129 J/gK, what was the initial temperature of the gold?
Use law of conservation of energy
Heat lost = Heat gained
q gained = - q lost
water is gaining heat and gold is losing heat
q = ms (Tf-Ti)
22.2*4.184(27.5-25)= -182*0.129*(27.5-x)
solve for x